Force and Moment Reactions Three cables intersect at B in order to stabilize a t

Question

Force and Moment Reactions

Three cables intersect at B in order to stabilize a telephone pole, as shown in the figure. The unit vector giving the direction of each cable is:
u1 = {(sqrt(6)/4)i -(sqrt(6)/4)j -(1/2)k}
u2 = {(2/sqrt(5))j -(1/ sqrt(5))k}
u3 = {-(1/sqrt(6))i +(1/sqrt(6))j -(2/sqrt(6))k}

H = 13 ft

F1 = 131 lb

F2 = 166 lb

F3 = 83.1 lb

Find:
(1) RAijk, the reaction force at A in {ijk} format
(2) MAijk, the resultant moment at A in {ijk} format
An example of {i,j,k} form: 22.4i-11.3j-70.3k

Explanation / Answer

>> Now, As, Forces, F1, F2 and F3 are acting along directions u1, u2 and u3

>> and, u1 = {(sqrt(6)/4)i -(sqrt(6)/4)j -(1/2)k},

u2 = {(2/sqrt(5))j -(1/ sqrt(5))k}, and

u3 = {-(1/sqrt(6))i +(1/sqrt(6))j -(2/sqrt(6))k}

and, F1 = 131 lb, F2 = 166 lb and F3 = 83.1 lb

Now, AB = H k = 13 k

>> Now, as Rb, i.e. Reaction at B = Resultant of F1, F2 and F3

=> Rb = 131(u1) + 166(u2) + 83.1(u3)

=> Rb = 131{(sqrt(6)/4)i -(sqrt(6)/4)j -(1/2)k} + 166{(2/sqrt(5))j -(1/ sqrt(5))k} + 83.1{-(1/sqrt(6))i +(1/sqrt(6))j -(2/sqrt(6))k}

=> Reaction at B, Rb = 194.77 i - 117.29 j - 207.59 k

>> As, In Beam AB,

Ra + Rb = 0, i.e There is no net force until and unless an external force is applied

=> Ra = - Rb

=> Ra = - 194.77 i + 117.29 j + 207.59 k

>> Now, Ma, i.e. Resultant moment at A

and, Ma = AB X Rb

=> Ma = 13 k X (194.77 i - 117.29 j - 207.59 k) = 2.532 j + 1.525 i {klb-ft}

=> Ma = Resultant Moment at A = 1.525 i + 2.532 j {klb - ft}      ......ANSWER........

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